Roman C. answered 08/09/15
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Masters of Education Graduate with Mathematics Expertise
Use the half-angle formulas: sin(θ/2) = √[(1-cos θ)/2] and cos(θ/2) = √[(1+cos θ)/2]
cos 18° = √[(1+cos 36°)/2] = √{[1+(1+√5)/4]/2} = √[(5+√5)/8] = √(10+2√5)/4
sin 9° = √[(1-cos 18°)/2] = √{[1-√(10+2√5) / 4]/2}
= √{[4-√(10+2√5)]/8} = √[8-2√(10+2√5)]/4
cos 9° = √[(1+cos 18°)/2] = √{[1+√(10+2√5) / 4]/2}
= √{[4+√(10+2√5)]/8} = √[8+2√(10+2√5)]/4